Factor f(x)=2x^{3}+(-3-2i)x^{2}+(-3-i)x+(2+i) into linear factors if 2+i is a zero of the function.

Question
Discrete math
asked 2021-03-11
Factor \(\displaystyle{f{{\left({x}\right)}}}={2}{x}^{{{3}}}+{\left(-{3}-{2}{i}\right)}{x}^{{{2}}}+{\left(-{3}-{i}\right)}{x}+{\left({2}+{i}\right)}\) into linear factors if \(\displaystyle{2}+{i}\) is a zero of the function.

Answers (1)

2021-03-12
Step 1
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Step 2
We Know that Synthetic Division is a method to divide a polynomial by a linear expression.
\(\displaystyle{f{{\left({x}\right)}}}={2}{x}^{{{3}}}+{\left(-{3}-{2}{i}\right)}{x}^{{{2}}}+{\left(-{3}-{i}\right)}{x}+{\left({2}+{i}\right)}\)
by synthetic division.
we have given 2+i is factor (zero) of function
\(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}{2}+{i}&{2}&-{3}-{2}{i}&-{3}-{i}&{2}+{i}\backslash&&{4}+{2}{i}&{2}+{i}&-{2}-{i}\backslash{h}{l}\in{e}-{1}&{2}&{1}&-{1}&{0}\backslash&&-{2}&{1}\backslash{h}{l}\in{e}&{2}&-{1}&{0}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\)
So, f(x) can be written in its linear form at.
\(\displaystyle{\left({x}-{\left({2}+{i}\right)}{\left({x}+{1}\right)}{\left({2}{x}-{1}\right)}={0}\right.}\)
[Here \(\displaystyle{2}+{i},-{1},{\frac{{{1}}}{{{2}}}}\) are zero of function]
0

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